1658.贵阳万科城E区二标公交车站工程施工方案 (1).docx

1658.贵阳万科城E区二标公交车站工程施工方案 (1).docx
仅供个人学习
反馈
文件类型:docx
资源大小:0.8 M
标准类别:施工组织设计
资源属性:
下载资源

施工组织设计下载简介

内容预览随机截取了部分,仅供参考,下载文档齐全完整

1658.贵阳万科城E区二标公交车站工程施工方案 (1).docx

Mmax=max[M1,M2]=max[0.059,0.198]=0.198kN·m

σ=Mmax/W=0.198×106/37500=5.272N/mm2≤[f]=15N/mm2

νmax=5ql4/(384EI)=5×4.116×2504/(384×10000×281250)=0.074mm

储罐水压试验施工方案 ν=0.074mm≤[ν]=L/250=250/250=1mm

q1=0.9×max[1.2(G1k+ (G2k+G3k)×h)+1.4Q1k,1.35(G1k +(G2k+G3k)×h)+1.4×0.7×Q1k]×b=0.9×max[1.2×(0.3+(24+1.1)×0.16)+1.4×2.5,1.35×(0.3+(24+1.1)×0.16)+1.4×0.7×2.5]×0.25=1.953kN/m

因此,q1静=0.9×1.2×(G1k +(G2k+G3k)×h)×b=0.9×1.2×(0.3+(24+1.1)×0.16)×0.25=1.165kN/m

q1活=0.9×1.4×Q1k×b=0.9×1.4×2.5×0.25=0.787kN/m

q2=0.9×1.2×G1k×b=0.9×1.2×0.3×0.25=0.081kN/m

p=0.9×1.4×Q1k=0.9×1.4×2.5=3.15kN

计算简图如下:

M1=0.125q1静L2+0.125q1活L2=0.125×1.165×12+0.125×0.787×12=0.244kN·m

M2=max[0.07q2L2+0.203pL,0.125q2L2+0.188pL]=max[0.07×0.081×12+0.203×3.15×1,0.125×0.081×12+0.188×3.15×1]=0.645kN·m

M3=max[q1L12/2,q2L12/2+pL1]=max[1.953×0.22/2,0.081×0.22/2+3.15×0.2]=0.632kN·m

Mmax=max[M1,M2,M3]=max[0.244,0.645,0.632]=0.645kN·m

σ=Mmax/W=0.645×106/42667=15.12N/mm2≤[f]=15.444N/mm2

V1=0.625q1静L+0.625q1活L=0.625×1.165×1+0.625×0.787×1=1.221kN

V2=0.625q2L+0.688p=0.625×0.081×1+0.688×3.15=2.218kN

V3=max[q1L1,q2L1+p]=max[1.953×0.2,0.081×0.2+3.15]=3.166kN

Vmax=max[V1,V2,V3]=max[1.221,2.218,3.166]=3.166kN

τmax=3Vmax/(2bh0)=3×3.166×1000/(2×40×80)=1.484N/mm2≤[τ]=1.782N/mm2

q=(γG(G1k +(G2k+G3k)×h))×b=(1×(0.3+(24+1.1)×0.16))×0.25=1.079kN/m

挠度,跨中νmax=0.521qL4/(100EI)=0.521×1.079×10004/(100×9350×170.667×104)=0.352mm≤[ν]=L/250=1000/250=4mm;

悬臂端νmax=ql14/(8EI)=1.079×2014/(8×9350×170.667×104)=0.014mm≤[ν]=2×l1/250=2×201/250=1.6mm

1、小梁最大支座反力计算

q1=0.9×max[1.2(G1k +(G2k+G3k)×h)+1.4Q1k,1.35(G1k +(G2k+G3k)×h)+1.4×0.7×Q1k]×b=0.9×max[1.2×(0.5+(24+1.1)×0.16)+1.4×1.5,1.35×(0.5+(24+1.1)×0.16)+1.4×0.7×1.5]×0.25=1.702kN/m

q1静=0.9×1.35×(G1k +(G2k+G3k)×h)×b=0.9×1.35×(0.5+(24+1.1)×0.16)×0.25=1.372kN/m

q1活=0.9×1.4×0.7×Q1k×b =0.9×1.4×0.7×1.5×0.25=0.331kN/m

q2=(γG(G1k +(G2k+G3k)×h))×b=(1×(0.5+(24+1.1)×0.16))×0.25=1.129kN/m

承载能力极限状态

按二等跨连续梁,Rmax=1.25q1L=1.25×1.702×1=2.128kN

按悬臂梁,R1=q1l1=1.702×0.2=0.34kN

R=max[Rmax,R1]=2.128kN;

正常使用极限状态

按二等跨连续梁,R'max=1.25q2L=1.25×1.129×1=1.411kN

按悬臂梁,R'1=q2l1=1.129×0.2=0.226kN

R'=max[R'max,R'1]=1.411kN;

计算简图如下:

主梁计算简图一

主梁弯矩图一(kN·m)

σ=Mmax/W=0.692×106/4490=154.034N/mm2≤[f]=205N/mm2

主梁剪力图一(kN)

τmax=2Vmax/A=2×3.352×1000/424=15.81N/mm2≤[τ]=125N/mm2

主梁变形图一(mm)

跨中νmax=0.862mm≤[ν]=1000/250=4mm

悬挑段νmax=0.049mm≤[ν]=2×250/250=2mm

5、支座反力计算

承载能力极限状态

支座反力依次为R1=7.288kN,R2=8.672kN,R3=8.672kN,R4=7.288kN

按上节计算可知,可调托座受力N=8.672kN≤[N]=30kN

顶部立柱段:l01=kμ1(hd+2a)=1×1.386×(1500+2×201)=2633mm

非顶部立柱段:l0=kμ2h =1×1.755×1800=3159mm

λ=max[l01,l0]/i=3159/15.9=198.679≤[λ]=210

2、立柱稳定性验算

q1=1×[1.2×(0.5+(24+1.1)×0.16)+1.4×0.9×1]×0.25 = 1.67kN/m

同上四~六步计算过程,可得:

R1=7.148kN,R2=8.505kN,R3=8.505kN,R4=7.148kN

l01=kμ1(hd+2a)=1.155×1.386×(1500+2×201)=3041.577mm

λ1=l01/i=3041.577/15.9=191.294

查表得,φ=0.197

不考虑风荷载:

N1 =Max[R1,R2,R3,R4]=Max[7.148,8.505,8.505,7.148]=8.505kN

f= N1/(ΦA)=8505/(0.197×424)=101.822N/mm2≤[f]=205N/mm2

Mw=1×γQφcωk×la×h2/10=1×1.4×0.9×0.076×1×1.82/10=0.031kN·m

N1w =Max[R1,R2,R3,R4]+Mw/lb=Max[7.148,8.505,8.505,7.148]+0.031/1=8.536kN

f= N1w/(φA)+ Mw/W=8536/(0.197×424)+0.031×106/4490=109.097N/mm2≤[f]=205N/mm2

非顶部立柱段:

l0=kμ2h =1.155×1.755×1800=3648.645mm

λ=l0/i=3648.645/15.9=229.475

查表得,φ1=0.139

不考虑风荷载:

N =Max[R1,R2,R3,R4]+1×γG×q×H=Max[7.148,8.505,8.505,7.148]+1×1.2×0.15×6.05=9.594kN

f=N/(φ1A)=9.594×103/(0.139×424)=162.787N/mm2≤[σ]=205N/mm2

Mw=1×γQφcωk×la×h2/10=1×1.4×0.9×0.076×1×1.82/10=0.031kN·m

大六角高强度螺栓连接分项工程质量管理技术、安全交底.doc Nw =Max[R1,R2,R3,R4]+1×γG×q×H+Mw/lb=Max[7.148,8.505,8.505,7.148]+1×1.2×0.15×6.05+0.031/1=9.625kN

f=Nw/(φ1A)+Mw/W=9.625×103/(0.139×424)+0.031×106/4490=170.217N/mm2≤[σ]=205N/mm2

H/B=6.05/1.6=3.781≥3

十、立柱地基基础验算

立柱底垫板的底面平均压力p=N/(mfA)=9.625/(0.9×0.1)=106.944kPa≤fak=140kPa

梁模板(扣件式,梁板立柱共用)计算书

2、《建筑施工扣件式钢管脚手架安全技术规范》JGJ 130-2011

GB/T 51250-2017标准下载 设计简图如下:

©版权声明
相关文章